Derivation of the Paschen curve law

ALPhA Laboratory Immersion

Arturo Dominguez

July 3, 2014

1 Objective

If a voltage di↵erential is supplied to a gas as shown in the setup (Figure 1), anelectric field is formed. If the electric field applied is strong enough, an avalancheprocess (the Townsend avalanche) is started which leads to the breakdown of thegas and the formation of plasma. This document describes the physics of thisprocess and derives the law (Paschen’s law) that predicts the voltage di↵erentialthat needs to be supplied in order to create the plasma.

2 Experimental Setup

In Figure 1 the experimental setup for the Paschen curve experiment is shown.A pair of parallel plate electrodes are placed inside a vessel that contains a gaswhich can be air but can also be a more pure gas, like He, Ar, Ne, etc. While thegeometry of the setup is irrelevant to the qualitative behavior of the breakdownvoltage, the simple 1D geometry results in a simpler comparison with theory.The variables that can be controlled are: the pressure of the gas, p, the distancebetween the electrodes, d, and the voltage between the electrodes, V , as well asthe gas contained in the vessel.

3 Qualitative description

As the voltage di↵erence is applied, the electric field will accelerate any freecharges that exist. Free electrons exist in the system due to random eventsfrom a variety of mechanisms including the triboelectric e↵ect or through as-tronomical particles traversing the vessel and ionizing neutral particles. If anelectron can gain more than the ionizing energy of the gas, U

I

(approximately14eV for Nitrogen), the electron can ionize the neutral particle and create anew free electron and a free ion. The new free electron can repeat the pro-cess and create a chain reaction called the Townsend Avalanche. The ion isaccelerated towards the cathode and as it hits it there is a chance that it will

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Figure 1: The standard DC discharge setup shows an electron ionizing a neutralparticle. When the process becomes self sustaining, the Townsend Avalanche isinitiated and a plasma is formed.

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free an electron from the cathode. This process is called secondary emissionand is responsible for the production of electrons that can sustain the plasma.Once a threshold condition is met, the secondary electrons su�ce to begin theTownsend Avalanche.

4 Derivation of Paschen’s Law

Quantitatively, the electron avalanche can be described using the rate of ion-ization per unit length, ↵. For example, if ↵ = 2cm�1 then within a cm ofelectron travel along the x-axis, there will, on average, be 2 collisions with neu-tral particles that result in ionizing events. This results in the following equationdescribing the increase of electron current density, �

e

(x):

d�e

(x) = �e

(x)↵dx, (1)

which integrates to:�e

(x) = �e

(0)e↵x. (2)

Since there is no current leaving or entering the vessel except through the elec-trodes, and there is no charge accumulation, the continuity equation results in:

�(0) = �(d), (3)

that is, the current density at the cathode (x = 0) is the same as that at theanode (x = d). Assuming a single ion species, Equation 3 can be rewritten as:

�e

(0) + �i

(0) = �e

(d) + �i

(d) (4)

�i

(0) = �e

(d)� �e

(0) + �i

(d). (5)

Using the fact that there is no input of ions from the anode, �i

(d) = 0. Usingthis constraint, and substituting Equation 2 into Equation 5 results in:

�i

(0) = �e

(0)(e↵d � 1). (6)

As an ion reaches the cathode, the probability of it releasing a secondary electronis given by the coe�cient �, therefore:

�e

(0) = ��i

(0). (7)

At the threshold point where the secondary emission sustains the plasma andthe Townsend Avalanche is formed, Equations 6 and 7 are balanced and �

i

(0)can be substituted, leading to:

�e

(0)

�

= �e

(0)(e↵d � 1) (8)

1

�

= e↵d � 1 (9)

↵d = ln (1 + 1/�). (10)

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Figure 2: The volume associated with an individual neutral gas particle is givenby vol = 1/n = ��.

Figure 3: The volume associated with an individual neutral gas particle is givenby vol = 1/n.

Equation 10 gives one constraint on the ionization coe�cient.While we’ve defined the role of ↵, we haven’s discussed a physical derivation ofwhat its value should be. In the following section we will tackle that.

As an electron is accelerated through the gas, the distance it travels, onaverage, before it collides with a neutral particle is given by the mean freepath (mfp), �. The volume occupied by a single neutral particle is given byvol = 1/n, where n is the neutral density, as shown in Figure 2. The face ofthe cylinder is the cross section of the collision between the neutral particle andthe electron, or � ⇡ ⇡(r

e

+ rn

)2 ⇡ ⇡r2n

where re

and rn

are the electron andneutral particle’s radii respectively. Therefore, the volume occupied per neutralis: vol = 1/n = ��. Since the neutral gas follows the ideal gas law, p = nk

B

T ,where p is the neutral pressure, T is the temperature of the neutrals and k

B

isthe Boltzmann constant, the mean free path can be rewritten as:

� =1

�n

=k

B

T

�p

(11)

� is the rate of collisions per length in the x-axis of electrons hitting neutralparticles. Note that if the electron is energetic enough, the collision will ionizethe particle, but this is not generally true.If you follow the flow of electrons at a position x0, one can define the currentdensity of free electrons, �

e free(x0+�) which is the current density of free elec-trons from x0 that have reached x0 +� without having collided with a neutral.

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In Figure 3 it is observed that as the electrons travel across the length, �, somewill collide with neutrals and are lost from the �

e free. The rate of collisions perunit length is, as was discussed earlier, �. Therefore, the di↵erential equationthat determines �

e free(x) is given as:

d�e free(x) = ��e free(x)

dx

�

(12)

�e free(x0 +�) = �e free(x0)e

��/� (13)

�e free(x0 +�)

�e free(x0)

= e��/� (14)

Note that the RHS of Equation 14 is independent of x0, hence, at any point inx, the probability that a free electron has traversed a distance of at least � isgiven by:

P (distance traveled � �) = e��/� (15)Now, let’s get to ↵. If every collision between an electron and a neutral resultedin an ionization, then ↵ = 1/�, but only the proportion of those electrons thathave enough energy to ionize the neutral particle, U

I

, will cause the ionization,therefore:

↵ =P (electrons with energy � U

I

)

�

(16)

Since the electrons gain energy by being accelerated down the electric field, E,then we can define �

I

using:

U

I

= eE�I

(17)

�

I

=U

I

eE

(18)

�

I

=U

I

d

eV

(19)

where E = V/d has been used. �I

is, therefore, the distance that an electronmust travel in the electric field in order to gain the necessary energy, U

I

, toionize the neutral upon collision. Therefore, Equation 20 can be rewritten as:

↵ =P (distance traveled � �

I

)

�

=e

��I/�

�

, (20)

where Equation 15 has been used. Multiplying d on both sides of Equation 20,we obtain:

↵d = de

��I/�

�

(21)

ln (1 + 1/�) =d

k

B

T/�p

e

(�UId/eV )/(kBT/�p) (22)

ln (1 + 1/�) =

✓�

k

B

T

◆(pd)e(�UI�/ekBT )(pd/V )) (23)

ln (ln (1 + 1/�))� ln (Apd) = �BpdV

(24)

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Figure 4: Equation 25 using A = 1.5mTorr�1m�1, B = 36V/(mTorr⇥m) and� = 0.02

where A ⌘ �/kB

T and B ⌘ UI

�/ek

B

T and we have used Equations 10, 11 and19. Finally, since V is really the voltage right at the point of breakdown, wecan change its notation to V

BD

(breakdown voltage) and put it on the LHS:

V

BD

=Bpd

ln (Apd)� ln (ln (1 + 1/�)) (25)

This is the final form of Paschen’s Law. If we take pd as the abscissa (x� axis)and V

BD

as the ordinate (y � axis) then the range (as we will see) is pd =(ln (1 + 1/�)/A,1).For A = 1.5mTorr�1m�1, B = 36V/(mTorr ⇥ m) and � = 0.02 which

are typical values in air at room temperature using stainless steel electrodes,Equation 25 leads to the plot shown in Figure 4.Note the existence of a minimum in the curve. The pd and V

BD

at the minimumcan be found using the fact that at the critical value, dVBD

d(pd) = 0:

dV

BD

d(pd|min)=

BD � 1pd|min (Bpd|min)

D

2= 0 (26)

B(D � 1) = 0 (27)D = ln (ln (1 + 1/�))� ln (Apd|min) = 1 (28)

pd|min =ln (1 + 1/�)

A

e (29)

V

BD min =B ln (1 + 1/�)

A

e (30)

where D ⌘ ln (ln (1 + 1/�)) � ln (Apd|min) is the denominator of Equation 25and e in Equations 29 and 30 is Euler’s number.

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Using Equations 10 and 11, Equation 29 can be rewritten as:

↵|min =e

�1

�

(31)

Or, comparing it to Equation 20, at the minimum: �I

= �. This makes senseintuitively, it’s taking the electrons the mfp to accelerate just enough to ionizethe neutrals. If you were to increase �

I

, a lot of the collisions would not leadto ionization, if you increase � then the rate of collisions would drop. (from thedenominator in Equation 20).We also see that when D = 0, V

BD

! 1, this occurs when:

ln (ln (1 + 1/�))� ln (Apd|inf) = 0 (32)ln (1 + 1/�) = Apd|inf (33)

↵|inf =1

�

(34)

This result also makes sense, since the rate of ionizing collisions can’t be greaterthan the rate of collisions (1/�). This sets the lower limit on pd as pd >ln (1 + 1/�)/A as discussed before.

5 Discussion of the minimum

The conditions for the minimum are shown in Equations 29 and 30. Its existencecan be understood by looking at the extremes: if pd is too big, since � ⇠ 1/p,� is very small, therefore the electrons will collide too much and won’t acquirethe ionizing energy U

I

necessary. On the other hand, if pd is too small, that is� is big compared to the distance between the electrodes d, the electrons willcollide with the anode before they have a chance of ionizing the gas.There are many implications of Paschen’s law, e.g., when constructing a fluo-rescent light bulb, where the distance between the electrodes d, is set by theapplication, the pressure is set so as to be close to the minimum in order tominimize the necessary voltage needed, leading to lower power consumption.The typical pressure inside a fluorescent bulb is ⇡ 2Torr..Another consequence of Paschen’s law is that if we’re dealing with atmosphericplasmas (p ⇡ 760Torr) the distance between electrodes for the minimum con-dition is calculated (using the same conditions as in Figure 4) to be ⇠ 10µm.This leads to the label of microplasmas for these type of discharges and theirconsequential di�culty in attaining experimentally.

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